A174592 - OEIS

%I #30 Sep 08 2022 08:45:51

%S 2,46,658,9182,127906,1781518,24813362,345605566,4813664578,

%T 67045698542,933826115026,13006519911838,181157452650722,

%U 2523197817198286,35143611988125298,489487370016555902,6817679568243657346

%N Numbers n such that n^2 + 2*(n+2)^2 is a square.

%C The equation n^2 + 2*(n+2)^2 = y^2 is transformed via x=3n+4 into the Diophantine equation x^2 - 3*y^2 = -8, and by division through 4 to (x/2)^2 - 3*(y/2)^2 = -2. Setting xbar = x/2 and ybar = y/2, the fundamental solution to xbar^2 - 3*ybar^2 = -2 is xbar = ybar = 1, and the general solution is given by multiplying (1+sqrt(3))*(u+sqrt(3)*v)^j, j=1,2,3,4,... where (u,v) = (A001075(j), A001353(j)). Expanding this product, isolating the square root., etc., and discarding the solutions that are associated with non-integer n generates the series of all solutions. - _R. J. Mathar_, May 02 2010

%C Also numbers n such that the sum of the four pentagonal numbers starting at index n is equal to the sum of four consecutive triangular numbers. - _Colin Barker_, Dec 19 2014

%H Vincenzo Librandi, <a href="/A174592/b174592.txt">Table of n, a(n) for n = 1..890</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).

%F From _Bruno Berselli_, Sep 07 2011:  (Start)

%F G.f.: 2*x*(1+8*x-x^2)/((1-x)*(1-14*x+x^2)).

%F a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).

%F a(n) = -4/3 + ((sqrt(3)+1)^(4n-1) - (sqrt(3)-1)^(4n-1))/(3*2^(2n-1)).  (End)

%t eq = Simplify[n^2 + 2*(n+2)^2 == y^2 /. n -> (x - 4)/3]; r = Reduce[x >= 0 && y >= 0 && eq, x, Integers] /. C[1] -> k; xx[k_] = x /. ToRules[r[[-1, -1]]]; Select[Table[Simplify[(xx[k] - 4)/3], {k, 1, 34}], IntegerQ] (* _Jean-François Alcover_, Sep 06 2011, after _R. J. Mathar_ *)

%t CoefficientList[Series[2 (1 + 8 x - x^2)/((1 - x) (1- 14 x + x^2)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Sep 24 2014 °)

%o (Magma) [n: n in [0..70000000] | IsSquare(3*n^2+8*n+8)];

%o (Magma) I:=[2,46,658]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..17]];  // _Bruno Berselli_, Sep 07 2011

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Apr 11 2010

%E More terms from _R. J. Mathar_, May 02 2010